GCE
Physics A
Unit H156/02: Depth in physics
Advanced Subsidiary GCE
Mark Scheme for June 2018
Oxford Cambridge and RSA Examinations
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2018
H156/02 Mark scheme June 2018
Here are the subject specific instructions for this question paper. CATEGORISATION OF MARKS
The marking schemes categorise marks on the MACB scheme. B marks These are awarded as independent marks, which do not depend on other marks. For a B-mark to be scored, the point to
which it refers must be seen specifically in the candidate’s answers.
M marks These are method marks upon which A-marks (accuracy marks) later depend. For an M-mark to be scored, the point to which it refers must be seen in the candidate’s answers. If a candidate fails to score a particular M-mark, then none of the dependent A-marks can be scored.
C marks These are compensatory method marks which can be scored even if the points to which they refer are not written down by the candidate, providing subsequent working gives evidence that they must have known it. For example, if an equation carries a C-mark and the candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then the C-mark is given.
A marks These are accuracy or answer marks, which either depend on an M-mark, or allow a C-mark to be scored.
SIGNIFICANT FIGURES If the data given in a question is to 2 sf, then allow to 2 or more significant figures. If an answer is given to fewer than 2 sf, then penalise once only in the entire paper. Any exception to this rule will be mentioned in the Additional Guidance.
3
H156/02 Mark scheme June 2018
Annotations available in RM Assessor Annotation Meaning
Correct response Used to indicate the point at which a mark has been awarded (one tick per mark awarded).
Incorrect response Used to indicate an incorrect answer or a point where a mark is lost.
AE Arithmetic error Do not allow the mark where the error occurs. Then follow through the working/calculation giving full subsequent ECF if there are no further errors.
BOD Benefit of doubt given Used to indicate a mark awarded where the candidate provides an answer that is not totally satisfactory, but the examiner feels that sufficient work has been done.
BP Blank page Use BP on additional page(s) to show that there is no additional work provided by the candidates.
CON Contradiction No mark can be awarded if the candidate contradicts himself or herself in the same response.
ECF Error carried forward Used in numerical answers only, unless specified otherwise in the mark scheme. Answers to later sections of numerical questions may be awarded up to full credit provided they are consistent with earlier incorrect answers. Within a question, ECF can be given for AE, TE and POT errors but not for XP.
L1 Level 1 L1 is used to show 2 marks awarded and L1^ is used to show 1 mark awarded.
L2 Level 2 L2 is used to show 4 marks awarded and L2^ is used to show 3 marks awarded.
L3 Level 3 L3 is used to show 6 marks awarded and L3^ is used to show 5 marks awarded.
POT Power of 10 error This is usually linked to conversion of SI prefixes. Do not allow the mark where the error occurs. Then follow through the working/calculation giving ECF for subsequent marks if there are no further errors.
SEEN Seen To indicate working/text has been seen by the examiner.
SF Error in number of significant figures
Where more SFs are given than is justified by the question, do not penalise. Fewer significant figures than necessary will be considered within the mark scheme. Penalised only once in the paper.
TE Transcription error This error is when there is incorrect transcription of the correct data from the question, graphical read-off, formulae booklet or a previous answer. Do not allow the relevant mark and then follow through the working giving ECF for subsequent marks.
XP Wrong physics or equation
Used in numerical answers only, unless otherwise specified in the mark scheme. Use of an incorrect equation is wrong physics even if it happens to lead to the correct answer.
^ Omission Used to indicate where more is needed for a mark to be awarded (what is written is not wrong but not enough).
4
H156/02 Mark scheme June 2018
Abbreviations, annotations and conventions used in the detailed Mark Scheme (to include abbreviations and subject-specific conventions).
Annotation Meaning
/ alternative and acceptable answers for the same marking point
Reject Answers which are not worthy of credit
Not Answers which are not worthy of credit
Ignore Statements which are irrelevant
Allow Answers that can be accepted
( ) Words which are not essential to gain credit
___ Underlined words must be present in answer to score a mark
ECF Error carried forward
AW Alternative wording
ORA Or reverse argument
5
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 1 (a) Deceleration is indicated by the negative gradient/slope
(between t =1.5 to t =4.5) A straight line/constant (negative) gradient (between t =1.5 to t =4.5) (indicates constant deceleration)
B1
B1
Constant negative gradient scores two marks
(b) area under graph = displacement or distance 12 × 3 × 14 = 21 (m) 1 (m)
C1
M1
A1
ALLOW 𝑠𝑠 = (𝑢𝑢+𝑣𝑣)2
𝑡𝑡 or 𝑠𝑠 = 𝑢𝑢𝑡𝑡 + 12𝑎𝑎𝑡𝑡
2 and a=14/3 (14+0)
2× 3 OR 7 × 3
Allow ECF for mis read of t or v Do not accept t = 4.5 Ignore “-“ sign
Total 5
6
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 2 Level 3 (5–6 marks)
Clear diagrams and procedure and measurements and analysis There is a well-developed line of reasoning which is clear and logically structured. The information presented is relevant and substantiated. Level 2 (3–4 marks) A diagram, some procedure, some measurements and some analysis. There is a line of reasoning presented with some structure. The information presented is in the most-part relevant and supported by some evidence. Level 1 (1–2 marks) Limited procedure and limited measurements or limited analysis There is an attempt at a logical structure with a line of reasoning. The information is in the most part relevant. 0 marks No response or no response worthy of credit.
B1 x6
Indicative scientific points may include: Diagram and procedure
• labelled diagram • correct circuit diagram • description of procedure • use of cushion in case load falls • repeats experiment.
Measurements
• use of balance to measure load • use of ruler to measure height • use stopwatch to measure time • use of ammeter to measure current • use of voltmeter to measure p.d.
Analysis
• equation to determine input power/energy (IV/IVt)
• equation to determine output power/energy (mgh/t or mgh)
• equation to determine efficiency • use of gradient of appropriate graph
Total 6
7
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 3 (a) h measured with a metre rule/ruler
(electronic) timer/data logger (started and stopped electronically) Method to start timer (and release ball), e.g. electromagnet or light gate to start timer Method to stop timer, e.g. trap door, second light gate
B1
B1
B1
B1
Allow metre stick, tape measure Not stopwatch Allow one mark for use light gates without reference to timer/starting/stopping
(b) 0.185 (s2)
B1
(c) (i) Plots one missing plot to less than a half small square Draws straight line of best fit
B1
B1
Allow ECF from (b) Allow ECF Expect to be balance of points about line of best-fit. Judge straightness by eye. Not thick lines, multiple lines
(ii) Determines gradient correctly and gradient in the range 0.210 to 0.225
B1
Ignore significant figures.
(d) (i) Evidence of use of 𝑠𝑠 = 𝑢𝑢𝑡𝑡 + 12𝑎𝑎𝑡𝑡
2 (and u = 0)
Manipulation leading to hg
t
=
22
B1
(ii) 𝒈𝒈 = 𝟐𝟐(c)ii
and given to 2 or 3 s.f. B1 Note: Possible ecf from (c)(ii) Ignore rounding
Total 10
8
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 4 (a) (weight of plank =) 50 x 9.81 or 490.5 OR
uses a distance of 0.7m to calculate clockwise moment (anticlockwise moment =) Tsin30° × 1.5 OR 0.75T (clockwise moment =) 490.5 × 0.7 = 343 (N m) Tsin30° × 1.5 = 343 OR Tsin30° = 229 T = 457.8 (N)
C1
C1
C1
C1
A0
Allow Tcos60° × 1.5 Allow 344, Allow 458.6,
(b) 𝑥𝑥 = 𝑇𝑇𝑇𝑇𝐸𝐸𝐸𝐸
𝑥𝑥 = 460 × 1.73
210 × 109 × 11 × 10−6
x = 3.45 × 10−4 (m)
C1
C1
A1
Note x must be the subject Allow alternative methods e.g. determines stress (4.18 × 107 Pa) C1 determines strain (1.99 × 10-4) C1 determines x Allow 3.4, 3.5, 3.43, 3.44 Allow 2 marks for 3.45 × 10𝑛𝑛
Total 7
9
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 5 (a) (Resultant) force is (directly) proportional to the rate of
change of momentum.
B1
Allow equation with symbols defined. Allow equal for proportional Ignore reference to direction Ignore F=ma
(b) (i) 0.868 or 0.87 B1
Allow – 0.868 or – 0.87
(ii) change in momentum = 0.5 × 17 × 0.18 or 1.53 (N s) Momentum of tennis ball = 1.53 – 0.868 = 0.662 v = 0.662/0.062 = 10.6 or 10.7 (m s-1) = 11 (m s-1)
C1
C1
A1
ECF from (b)(i) Or 0.062(v- -14) = 1.53 Or v = 24.7 -14 = 10.7 Allow 1 mark for 24.7 (m s-1) or 38.7 (m s-1)
(iii) For an elastic collision, kinetic energy/KE is conserved speeds are different (so for the same mass KE is different)
B1
B1
Allow speed of approach = speed of separation Allow correct calculations of KE for both speeds Ignore reference to the ball heating up
Total 7
10
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 6 (a) X (filament) lamp
Y (fixed)(ohmic) resistor
B1 Allow ptc thermistor / heater element Not metallic conductor
(b) Ix = 0.5 A and IY = 0.36 A OR I= 0.86 A 9.6V – 7.2V or 2.4 V 𝑟𝑟 = 9.6−7.2
0.86= 2.8 Ω
C1
C1
A1
Allow Alternative correct methods 2.79 Ω
(c) (i) Micrometer Repeat readings in different directions/along wire/different wires and average
B1
B1
Allow calliper Not vernier scale
(ii) 𝐸𝐸 = 𝜋𝜋×0.12×10−32
4= 1.13 × 10−8 OR 𝜌𝜌 = 1.86×𝐴𝐴
21
𝜌𝜌 =17 × 1.86 × 1.1 × 10−8
21
𝜌𝜌 = 1.7 × 10−8 (Ω m)
C1
C1
A1
Note ρ must be the subject
Allow 2 marks for 1.0 × 10-9 (factor of 17 omitted) Allow 2 marks for 6.8 × 10-8 (diameter used instead of radius) Allow 2 marks for 0.017 (POT omitted)
(iii) 0.121
or 0.021.86
or 0.010.12
0.121
+ 0.021.86
+ 2 × 0.010.12
× 100 = 18 (.2)%
C1
A1
Allow max/min methods ρmax = 2.03 × 10−8 and ρmin = 1.41 × 10−8 (B1) ∆𝜌𝜌𝜌𝜌𝑥𝑥100 (B1)
Allow 17.8% Do not penalise significant figures Allow 1 mark for 9.88% Allow 20% with evidence of working
Total 11
11
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 7 (a) Level 3 (5–6 marks)
Clear explanation of observations and correct determination of frequency. There is a well-developed line of reasoning which is clear and logically structured. The information presented is relevant and substantiated. Level 2 (3–4 marks) Clear explanation of observations or correct method to determine the frequency or some explanation of observations and some method for the determination of the frequency There is a line of reasoning presented with some structure. The information presented is in the most-part relevant and supported by some evidence. Level 1 (1–2 marks) Has limited explanation of observations or limited evidence of method to determine the frequency There is an attempt at a logical structure with a line of reasoning. The information is in the most part relevant. 0 marks No response or no response worthy of credit.
B1 x6
Indicative scientific points may include: Explanation of observations • Metal sheet reflects microwaves • Idea/description of superposition • Constructive/destructive interference • Standing wave pattern between T and
plate • Maxima are antinodes and and minima are
nodes. • Phase difference at nodes and antinodes • Distance between successive
maxima/minima is λ/2 • Distance between adjacent regions of
maximum and minimum intensities is λ/4
Determination of frequency • 𝑓𝑓 = 𝑣𝑣
𝜆𝜆
• λ = 4 ᵡ 72 mm = 288 mm • 𝑓𝑓 = 3×108
288×10−3= 1.04 × 109Hz
(b) Microwaves from T are transverse/polarised wtte At 0° or 180° the grille blocks (all) the (polarised) waves and at 90°the grille allows all the microwaves to pass.
B1
B1
Allow E field perpendicular to direction of motion Allow explanation in terms of 𝐼𝐼 = 𝐼𝐼0cos2𝜃𝜃
Total 8
12
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 8 (a) (i) 1.36
1.97 × 108
B1
B1
Not 1.3 or 1.4 Not 1.9 or 2.0
(ii) 5.2×10−7
1.52= 3.4(2) × 10-7 (m) B1 Allow 3.41 × 10-7 (m)
Not ECF from (a)(i) (b) (i) sin𝜃𝜃 =
sin371.52
(= 0.39593) 𝜃𝜃 = 23(.3)°
C1
A1
(ii) Ray in glass bends towards normal and ray in ethanol bends away from normal but at a smaller angle than 37ᵒ Rays are straight by eye
B1
Note Ray should not be parallel to incoming ray. Not angle of refraction is zero in glass
Total 6
13
H156/02 Mark scheme June 2018
Question Answer Marks Guidance 9 (a) (i) Energy of a photon
B1 Ignore h is Planck constant and f is frequency
(ii) Minimum energy required to remove/emit (a single) electron from the metal surface
B1 Ignore ‘it is work function’ Ignore photoelectric effect
(b) (i) 4.1 eV = = 4.1 × 1.6 × 10-19 or 6.56 × 10-19 J OR 𝐸𝐸𝑘𝑘 = 6.63 × 10−34 × 1.2 × 1015 − 𝜙𝜙 𝐸𝐸𝑘𝑘 = 6.63 × 10−34 × 1.2 × 1015 − 6.56 x 10−19 𝐸𝐸𝑘𝑘 = 1.39 × 10−19J
𝑣𝑣 = 2 × 1.39 × 10−19
9.11 × 10−31= 3.06 × 1011
5.536 × 105 m s-1
C1
C1
C1
C1
A0
Allow f0 = 9.9 × 1014 Hz Allow 𝐸𝐸𝑘𝑘 = 6.63 × 10−34 × (1.2 × 1015 − 9.9 x 1014) Allow 1.4 × 10-19 J 3.06 × 1011 scores three marks
(ii) Maximum energy is independent of intensity/(number of photons has increased but) energy of photon is the same/energy of a photon is only dependent on frequency/intensity affects the number of photons/electrons released only/frequency of photon has not changed No change in maximum speed
M1
A1
Not “Does not increase”
(c) 𝜆𝜆 =
ℎ𝑚𝑚𝑣𝑣
=6.63 × 10−34
9.11 × 10−31 × 5.5 × 105
= 1.3(2) × 10−9 (m)
C1
A1
Total 10
14
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